Chapter 10 - Geometry - 10.5 Volume and Surface Area - Exercise Set 10.5 - Page 657: 31

The surface area is\[186\text{ y}{{\text{d}}^{2}}\]and the volume is\[\text{148 y}{{\text{d}}^{3}}\].

There are two rectangular solids in the given figure. The dimensions that are the length, width, and height of the first rectangular solid is 5yd, 4yd, and 5yd, respectively. The dimensions of the second rectangular solid are 4yd, 4yd, and 3yd, respectively. To compute the surface area of the figure, firstly compute the surface area of the first rectangular solid and then compute the surface area of the second rectangular solid. Thirdly, compute the surface area of the figure by deducting the surface area of the unrequired rectangular part having dimensions as 3yd. and 4 yds. from the resultant of the sum of the resultant of the surface area of both the rectangular solids. Compute the surface area of the larger rectangular solid using the equation as shown below: \[\text{Surface area of the rectangular solid}=2\times \left[ \left( 5\text{ yd}\times 4\text{ yd} \right)+\left( 5\text{ yd}\times 5\text{ yd} \right)+\left( 4\text{ yd}\times 5\text{ yd} \right) \right]\] \[\begin{align} & =2\times \left[ \left( 20 \right)+\left( 25 \right)+\left( 20 \right) \right]\text{ y}{{\text{d}}^{2}} \\ & =2\times 65\text{ y}{{\text{d}}^{2}} \\ & =130\text{ y}{{\text{d}}^{2}} \end{align}\] Compute the surface area of the smaller rectangular solid using the equation as shown below: \[\text{Surface area of the rectangular solid}=2\times \left[ \left( 4\text{ yd}\times 4\text{ yd} \right)+\left( 4\text{ yd}\times 3\text{ yd} \right)+\left( 4\text{ yd}\times 3\text{ yd} \right) \right]\] \[\begin{align} & =2\times \left[ \left( 16 \right)+\left( 12 \right)+\left( 12 \right) \right]\text{ y}{{\text{d}}^{2}} \\ & =2\times 40\text{ y}{{\text{d}}^{2}} \\ & =80\text{ y}{{\text{d}}^{2}} \end{align}\] Now, compute the surface area of the unrequired part using the equation as shown below: \[\begin{align} & \text{Surface area of the unrequired part}=2\times \left( 3\times 4 \right)\text{y}{{\text{d}}^{2}} \\ & =24\text{ y}{{\text{d}}^{2}} \end{align}\] Compute the surface area of the figure using the equation as shown below: \[\begin{align} & \text{Surface area of the figure}=\text{Surface area of the larger rectangular solid}+\text{Surface area of the } \\ & \text{smaller rectangular solid}-\text{Surface area of the unrequired part} \\ & =\left( 130+80-24 \right)\text{ y}{{\text{d}}^{2}} \\ & =186\text{ y}{{\text{d}}^{2}} \end{align}\] To compute the volume of the figure, firstly compute the volume of the first rectangular solid and then compute the volume of the second rectangular solid. Finally, compute the sum of the resultant of the volume of both the rectangular solids. Compute the volume of the larger rectangular solid using the equation as shown below: \[\begin{align} & \text{Volume of rectangular solid}=\left( 5\times 4\times 5 \right)\text{ y}{{\text{d}}^{3}} \\ & =100\text{ y}{{\text{d}}^{3}} \end{align}\] Now, compute the volume of the smaller rectangular solid using the equation as shown below: \[\begin{align} & \text{Volume of rectangular solid}=lwh \\ & =\left( 4\times 4\times 3 \right)\text{ y}{{\text{d}}^{3}} \\ & =48\text{ y}{{\text{d}}^{3}} \end{align}\] Compute the volume of the figure using the equation as shown below: \[\begin{align} & \text{Volume of the figure}=\text{Volume of larger rectangular solid}+\text{Volume of } \\ & \text{smaller rectangular solid} \\ & =\left( 100+48 \right)\text{ y}{{\text{d}}^{3}} \\ & =148\text{ y}{{\text{d}}^{3}} \end{align}\]